The set of all x values for which |f(x) + g(x)| < |f(x)| + |g(x)| is true if f(x) = x – 3 and g(x) = 4 – x is given by
Answer & explanation
Correct answer: option 3
$|f(x) + g(x)| < |f(x)| + |g(x)|$
$⇒|x-3+4-x| < |x-3|+|4-x|$
$1<|x-3|+|4-x|$
for equation if $x∈(3,4)$ then inequality doesn't hold
so R –[3, 4]